distribution comparison for large number of data sets - r

Question

I have data named "data" like this:
CENTRE_Blinded val_list
1 1104 c(-13, -1, 0, 28, -88, 28, -1, -6, -5, -58, 28, 28, 28, 28, 2, 0, 28, -26, 28, 28, 2, 28, 28, -2, -29, 0, 28, -34, -6, 0, 28, 1, 0, 0, -1, 28, 28, 0, 28, 6, 28, 0, 28, 28, 28, 0, -2, -6, -1, 4, 6, 1, -16, -7, 2, 3, 7, 0, 1, 11, 0, 1, -6, -5, 0, 3, 8, 7, 0, 0, 6, -6, 2, 36, -8, 0, -7, -7, -1, -1, -1, 7, -3, 7, 2)
2 1204 c(2, -9, 28, 28, -2, 1, -3, -1, 0, 28, 28, 28, 28, 28, 28, 3, 10, -5, -8, 9, -8, 0, 13, 0, -1, 2, -1, 0, 6, 1, 0, -7, 6, -6, 1)
3 1403 c(0, 2, 0, 2, 28, 0, -1, -35, -36, 2, 1, 1, 28, 28, 28, 0, 0, 28, -7, -35, 28, -3, -18, 28, 28, 28, -5, 0, 28, -2, 4, 5, 0, 56, 1, 0, 1, -7, -20, 0, 0, -3, 0, 1, 3, 0, 4, -2, 42, -13, 7, 10, 7, 56, 0, -5, 10, 56, 8, 56, 84, -4, 1, 0, -14, -7, -1, -48, -6, -3, 0, 7)
4 1110 c(0, 1, 0, -3, 28, 28, 0, -5, 0, 9, 15, 56, -11, -1, -7)
The first column containts ID of the centre, and the second contains list of values.
I wanto to build empirical distributions from these values for each centre and compare them pairwaise using e.g. kolmogorov-smirnov test (ks.test in R).
That way I would get N x N matrix of p-values.
My question is how to do it and preserve IDs of centres for each k-s test.
My try was:
val_list_temp = as.list(data, by = "CENTRE_Blinded"))
val_list = val_list_temp[[2]]
names(val_list) = val_list_temp[[1]]
Here I have IDs for each centre, but when I use expand.grid I don't no how to store them any more:
val_table = as.data.table(expand.grid(val_list, val_list))
ks_tests = apply(X = val_table, 1, function(x) ks.test(unlist(x[1]),unlist(x[2])))
Besides, I how to put it in matrix later, to do some visualisation?
P.S. Maybe there is a better way to do this than using kolmogorov-smirnov test and making matrix of p-values?

Answer 1

Is this the type of solution you were looking for? It loops 6 times (4 choose 2) and then makes the matrix symmetrical.
#Data you provided
A1104 <- c(-13, -1, 0, 28, -88, 28, -1, -6, -5, -58, 28, 28, 28, 28, 2, 0, 28, -26, 28, 28, 2,
28,28, -2, -29, 0, 28, -34, -6, 0, 28, 1, 0, 0, -1, 28, 28, 0, 28, 6, 28, 0, 28,
28, 28, 0, -2, -6, -1, 4, 6, 1, -16, -7, 2, 3, 7, 0, 1, 11, 0, 1, -6, -5, 0, 3, 8, 7,
0, 0, 6, -6, 2, 36, -8, 0, -7, -7, -1, -1, -1, 7, -3, 7, 2)
A1204 <- c(2, -9, 28, 28, -2, 1, -3, -1, 0, 28, 28, 28, 28, 28, 28, 3, 10, -5, -8, 9, -8, 0, 13,
0, -1, 2, -1, 0, 6, 1, 0, -7, 6, -6, 1)
A1403 <- c(0, 2, 0, 2, 28, 0, -1, -35, -36, 2, 1, 1, 28, 28, 28, 0, 0, 28, -7, -35, 28, -3, -18,
28, 28, 28, -5, 0, 28, -2, 4, 5, 0, 56, 1, 0, 1, -7, -20, 0, 0, -3, 0, 1, 3, 0, 4, -2,
42, -13, 7, 10, 7, 56, 0, -5, 10, 56, 8, 56, 84, -4, 1, 0, -14, -7, -1, -48, -6, -3, 0, 7)
A1110 <- c(0, 1, 0, -3, 28, 28, 0, -5, 0, 9, 15, 56, -11, -1, -7)
data <- list(A1104,A1204,A1403,A1110)
names(data) <- c("A1104","A1204","A1403","A1110")
#Provide combinations that we will want to compare
VarComb = combn(1:length(data), 2)
#Create matrix of all zeros and then populate with ks.test
Result.Matrix <- matrix(0,nrow=length(data),ncol=length(data))
for(i in 1:(dim(VarComb)[2])){
Result.Matrix[VarComb[1,i],VarComb[2,i]] <- ks.test(data[[VarComb[1,i]]],data[[VarComb[2,i]]])$p.value
}
Result.Matrix <- Result.Matrix+t(Result.Matrix)
row.names(Result.Matrix) <- names(data)
colnames(Result.Matrix) <- names(data)
Result.Matrix